SOLUTION: I need to know how to solve this equation. (x+18)(x-12)(x+3)>0. If anyone can help, I would appreciate it, thanks in advance.

Question 469683: I need to know how to solve this equation. (x+18)(x-12)(x+3)>0. If anyone can help, I would appreciate it, thanks in advance.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First, find the zeros of .

(x+18)(x-12)(x+3)=0

x+18=0 or x-12=0 or x+3=0

x=-18, x=12, or x=-3

In ascending order, the zeros are: -18,-3,12

Now in the interval

, (x+18)(x-12)(x+3) is negative. Simply plug in any negative number less than -18 to see this (eg: plug in x=-20 to get (-20+18)(-20-12)(-20+3)=-1088)

In the interval

, (x+18)(x-12)(x+3) is positive. Simply plug in any number in the interval to see this (eg: plug in x=-5 to get (-5+18)(-5-12)(-5+3)=442)

In the interval

, (x+18)(x-12)(x+3) is negative. Simply plug in any number in the interval to see this (eg: plug in x=0 to get (0+18)(0-12)(0+3)=-648)

Finally, in the interval

, (x+18)(x-12)(x+3) is positive. Simply plug in any number greater than 12 to see this (eg: plug in x=50 to get (50+18)(50-12)(50+3)=136952)

So the following intervals make (x+18)(x-12)(x+3) positive:

and

So combine them with a union symbol to get the final answer:

Here's a graph to visually confirm this

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